Proof. We make the antithesis that fff does not vanish identically. Then there exists a pointx0subscriptx0x_{0} of the open interval(a,b)ab(a,\,b) such that f⁢(x0)≠0fsubscriptx00f(x_{0})\neq 0; for example f⁢(x0)>0fsubscriptx00f(x_{0})>0. The continuity of fffimplies that there are the numbersαα\alpha and ββ\beta such that a<α<x0<β<baαsubscriptx0βba<\alpha<x_{0}<\beta<b and f⁢(x)>0fx0f(x)>0 for all
x∈[α,β]xαβx\in[\alpha,\,\beta]. Now the function φ0subscriptφ0\varphi_{0} defined by